Q. 134.7( 6 Votes )

# Glycerine flows s

Answer :

The Poiseuille’s law states the relation:

Where,

‘p’ is the difference in the pressure between the two ends.

‘r’ is the radius of the tube.

‘ɳ’ is the viscosity of the fluid.

‘l’ is the horizontal length of the tube.

Given,

Length of the tube, l = 1.5 m

Radius of the tube, r = 1 cm=0.01 m

Mass of glycerine flowing per second, M = 4.0 × 10^{-3} kg/s

Density of the glycerine, ρ = 1.3 × 10^{3} kg/m^{3}

Viscosity of the glycerine, ɳ = 0.83 Pa s

Volume of glycerine flowing per second is given as,

⇒V=

⇒ V=3.08 × 10^{-6} m^{3}/sec

Applying Poiseuille’s law we get the relation,

⇒

⇒ p=9.8 × 10^{2} Pa

To check whether the assumption of laminar flow in the tube is correct we use the Reynolds’s number relation given as,

Where,

and if the Reynolds number is **less than 2000**, the flow is laminar.

Where,

‘ρ’ is the density of the fluid.

‘V’ is the volume of fluid flowing per sec

‘d’ is the diameter of the pipe

‘ɳ’ is the viscosity of the fluid

Thus,

⇒

⇒ R =0.3

Since, the Reynolds’s number is less than 2000. Hence, the flow is laminar.

__The pressure difference between the two ends of the tube is__ __9.8 × 10__^{2}__Pa__

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